Oxygen and Carbon

If you start with 6 grams of carbon and 16 grams of oxygen and they react completly, how many grams of carbon dioxide will be produced? a. 22 b. 16 c. 10 d. 6

If you start with 3 grams of carbon and 8 grams of oxygen and they react completly, how many grams of carbon dioxide will be produced? a. 11 b. 5 c. 8 d. 3

Answer #1

I’ll put it in anyways. Cheater!

Answer #2

Um no I am not cheating. The teachers said we can look up the answers if we wanted to so don’t put your opinion in if it’s not going to help.

Answer #3

Are you cheating on a Chemistry Test?!

:O) Tsk tsk tsk…

Answer #4

Ye gods. You could at least attempt to disguise the fact that it’s a homework problem.

Answer #5

1)a) 2)a)

(lol, I have no idea at all. I suck at chemestry. That was the most logical answer…)

Answer #6

Instead of yelling at you I’ll tell you what to do.

If you ever see a question like this the first thing you need to do is convert to moles. You know how many grams of carbon and oxygen you have and if you have a periodic table you can find out how many grams are in a mole. I think its something like 12 in carbon and 15.9 in oxygen. So do the math and find out how many moles of each you have. HINT: Both of these numbers should be less than one because a mole is a big amount of a substance. Like you could hold a mole of carbon in your hand.

Now you need to find out what the reaction is. We know that O + C are turning into CO2. But you know how sometimes some elements have the number 2 after them. The way you remember these elements is by remembering HONClBrIF. Notice that the element oxygen (O) is included. This means the reaction is O2 + C -> CO2 So for every 2 moles of oxygen you have you can make one mole of c02. And for every mole of C you can make one mole of CO2. This might be a limiting agent problem so figure out how much co2 you can make by reacting all the o2, and then how much you can make using the c. The answer with less moles should be right. In the end convert the number of moles back to grams

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